NCERT Solutions for Class 8 Maths - (Ex: 6.3 - 6.4) - Squares and Square Roots

Exercise 6.3 

Question 1: What could be the possible ‘one’s’ digits of the square root of each of the following numbers: 

(i) 9801 

(ii) 99856 

(iii) 998001 

(iv) 657666025 

Answer 1: 

Since, Unit’s digits of square of numbers are 0, 1, 4, 5, 6 and 9. Therefore, the possible unit’s digits of the given numbers are:

(i) 1

(ii) 6

(iii) 1

(iv) 5

Question 2: 

Without doing any calculation, find the numbers which are surely not perfect squares:

(i) 153 

(ii) 257

(iii) 408 

(iv) 441 

Answer 2:

Since, all perfect square numbers contain their unit’s place digits 0, 1, 4, 5, 6 and 9.

(i) But given number 153 has its unit digit 3. So it is not a perfect square number.

(ii) Given number 257 has its unit digit 7. So it is not a perfect square number.

(iii) Given number 408 has its unit digit 8. So it is not a perfect square number.

(iv) Given number 441 has its unit digit 1. So it would be a perfect square number

Question 3: 

Find the square roots of 100 and 169 by the method of repeated subtraction. 

Answer 3: 

By successive subtracting odd natural numbers from 100,

This successive subtraction is completed in 10 steps.

By successive subtracting odd natural numbers from 169,

This successive subtraction is completed in 13 steps.

Therefore

Question 4: Find the square roots of the following numbers by the Prime Factorization method:

(i) 729

(ii) 400

(iii)1764

[iv] 4096

[v] 7744

[vi] 9604

[vil] 5929

(viii) 9216

[ix] 529

[x] 8100

Answer:

Question 5: 

For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also, find the square root of the square number so obtained: 

(i) 252 

(ii) 180 

(iii) 1008 

(iv) 2028 

(v) 1458 

(vi) 768 

Answer 5:

(i) 

252 = 2x2x3x3x7

Here, prime factor 7 has no pair. Therefore 252 must be multiplied by 7 to make it a perfect square.

252x7 = 1764

and √1764 = 2 x 3 x 7 = 42

(ii) 

Here, prime factor 5 has no pair. Therefore 180 must be multiplied by 5 to make it a perfect square.

180x5 = 900

and √900 = 2 x 3 x 5 = 30

(iii)

1008 = 2x2x2x2x3x3x7

Here, prime factor 7 has no pair. Therefore 1008 must be multiplied by 7 to make it a perfect square.

1008x5 = 7056

and √7056 = 2 x 2x 3 x 7 = 84

(iv)

2028 = 2x2x3x13x13

Here, prime factor 3 has no pair. Therefore 2028 must be multiplied by 3 to make it a perfect square.

2028x3 = 6084

and √6084 = 2 x 2 x 3 x 3 x 13 x 13 = 78

(v) 1485 = 2 x 3 x3 x3 x3 x3

Here, prime factor 2 has no pair. Therefore 1458 must be multiplied by 2 to make it a perfect square.

1458x2 = 2916

and √2916 = 2 x 2 x 3 x 3 = 54

(vi)

768 = 2x2x2x2x2x2x2x3

Here, prime factor 3 has no pair. Therefore 768 must be multiplied by 3 to make it a perfect square.

768x3 = 2304

and √2304 = 2 x 2 x 2 x 2 x 3 = 48

Question 6: 

For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also, find the square root of the square number so obtained: 

(i) 252 

(ii) 2925 

(iii) 396 

(iv) 2645 

(v) 2800 

(vi) 1620 

Answer 6: 

(i)

252 =2x2x3x3x7

Here, prime factor 7 has no pair. Therefore 252 must be divided by 7 to make it a perfect square.

252/7 = 36

and √36 = 2 x 3 = 6

(ii) 

2925 = 3x3x5x5x13

Here, prime factor 13 has no pair. Therefore 2925 must be divided by 13 to make it a perfect square. 

180 x 5 = 900

and √900 = 2 x 3 x 5 = 30

(iii)

396 = 2x2x3x3x11

Here, prime factor 11 has no pain Therefore 396 must be divided by 11 to make it a perfect square.

396 / 11 = 36

and √36 = 2 x 3  = 6

(iv)

2645 = 5x23x23

Here, prime factor 5 has no pair. Therefore 2645 must be divided by 5 to make it a perfect square.

2645/5 = 529

and √529= 23  x 23 = 23

(v)

2800 =2x2x2x2x5x5x7

Here, prime factor 7 has no pair. Therefore 2800 must be divided by 7 to make it a perfect square.

2800 / 7 = 400

and √400 = 20  x 20 = 20

(vi)

1620 =2x2x3x3x3x3x5

Here, prime factor 5 has no pair. Therefore 1620 must be divided by 5 to make it a perfect square.

1620 /5 = 324

and √324 = 2  x 3 x 3 = 18

Question 7: 

The students of Class VIII of a school donated  2401 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class. 

Answer 7: 

Question 8: 

2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row. 

Answer 8: 

Here, Number of plants = 2025

Let the number of rows of planted plants be x

And each row contains number of plants = x

Question 9: 

Find the smallest square number that is divisible by each of the numbers 4, 9 and 10. 

Answer 9: 

L.C.M. of 4,9 and 10 is 180.

Prime factors of 180 =2 x 2 x 3 x 3 x 5

Here, prime factor 5 has no pair. Therefore 180 must be multiplied by 5 to make it a perfect square.

∴ 180 x 5 = 900

Hence, the smallest square number which is divisible by 4, 9 and 10 is 900.

Question 10: 

Find the smallest square number that is divisible by each of the numbers 8, 15 and 20. 

Answer 10: 

L.C.M. of 8,15 and 20 is 120.

Prime factors of 120 = 2 x 2 x 2 x 3 x 5

Here, prime factor 2, 3 and 5 has no pair. Therefore 120 must be multiplied by

2 x 3 x 5 to make it a perfect square,

∴120 x 2 x 3 x 5 = 3600

Hence, the smallest square number which is divisible by 8,15 and 20 is 3600.   

Exercise 6.4 

Question 1: 

Find the square roots of each of the following numbers by Division method: 

(i) 2304

(ii) 4489

(iii) 3481

(iv) 529

[v] 3249

(vi) 1369

(vii) 5776

(viii)7921

(ix) 576

(x) 1024

(xi) 3136

(xii) 900

Answer : 

Question 2: 

Find the number of digits in the square root of each of the following numbers (without any calculation): 

(i) 64 (ii) 144 (iii) 4489 (iv) 27225 (v) 390625 

Answer : 

(i) 

Here, 64 contains two digits which is even.

Therefore, number of digits in square root = n/2 = 2/2 = 1

(ii)

Here, 144 contains two digits which is odd

Therefore, number of digits in square root = 

(iii)

Here, 4489 contains two digits which is even

Therefore, number of digits in square root = n/2 = 4/2 = 2

(iv)
Here, 390625 contains two digits which is even

Therefore, number of digits in square root = n/2 = 6/2 = 3

Question 3: 

Find the square root of the following decimal numbers: 

(i) 2.56 

(ii) 7.29 

(iii) 51.84 

(iv) 42.25 

(v) 31.36 

Answer : 

Question 4: 

Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square so obtained: 

(i) 402 

(ii) 1989 

(iii) 3250 

(iv) 825 

(v) 4000 

Answer:

(i)

We know that, if we subtract the remainder from the number, we get a perfect square.

Here, we get remainder 2. Therefore 2 must be subtracted from 402 to get a perfect square.

∴ 402-2 =400

Hence, the square root of 400 is 20.

(ii)

We know that, if we subtract the remainder from the number, we get a perfect square.

Here, we get remainder 53. Therefore 53 must be subtracted from 1989 to get a perfect square.

∴ 1989- 53 = 1936

Hence, the square root of 1936 is 44

(iii)

We know that, if we subtract the remainder from the number, we get a perfect square.

Here, we get remainder 1. Therefore 1 must be subtracted from 3250 to get a perfect square.

∴ 3250- 1 = 3249

Hence, the square root of 3249 is 57

(iv)

We know that, if we subtract the remainder from the number, we get a perfect square.

Here, we get remainder 41. Therefore 41 must be subtracted from 825 to get a perfect square.

∴ 825-41 = 784

Hence, the square root of 784 is 28.

(v)

We know that, if we subtract the remainder from the number, we get a perfect square.

Here, we get remainder 31. Therefore 31 must be subtracted from 4000 to get a perfect square.

∴ 4000- 31 = 3969

Hence, the square root of 3969 is 63.

Question 5: 

Find the least number which must be added to each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square so obtained: 

(i) 525 

(ii) 1750 

(iii) 252 

(iv) 1825 

(v) 6412 

Answer : 

Question 6: 

Find the length of the side of a square whose area is 441 m²? 

Answer : 

Hence, the length of the side of a square 21 m.

Question 7: 

In a right triangle ABC,  B = 90 . 

(i) If AB = 6 cm, BC = 8 cm, find AC. 

(ii) If AC = 13 cm, BC = 5 cm, find AB. 

Answer 7: 

Question 8: 

A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and number of columns remain same. Find the minimum number of plants he needs more for this. 

Answer :

Here, plants = 1000

Since remainder is 39. Therefore 31² < 1000

Next perfect square number '32²  = 1024

Hence, number to be added = 1024 - 1000 = 24

= 1000 + 24 = 1024

Hence, the gardener required 24 more plants.

Question 9: 

There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the number of rows is equal to number of columns. How many children would be left out in this arrangement? 

Answer : 

Here, Number of children = 500

By getting the square root of this number, we get,

In each row, the number of children is 22.

And left out children are 16.